Complex Numbers

Introduction

The motivation for defining complex numbers comes from wanting to define ${\sqrt {-1}}$ in a way that would be compatible with the properties of exponents. Mathematicians accomplished this by defining a new "variable", i. Thus, we have the immediate property that $i^{2}=-1$, the same way $({\sqrt {3}})^{2}=3$. Now we can define the Complex Numbers to be all numbers of the form $A+Bi$ for real numbers $A$ and $B$. The number A is called the real part of $A+Bi$, while B is the imaginary part of $A+Bi$.

Properties

Before we start describing the properties that Complex Numbers satisfy, we have to define what it means for Complex Numbers to be equal,

1) If A + Bi and C + Di are two complex numbers, then A + Bi = C + Di if and only if A = C and B = D. For example, $1+3i\neq 2+6i$ since $1\neq 2$

2) If A + Bi and C + Di are two complex numbers, then we can add them together as follows: (A + Bi) + (C + Di) = (A + B) + (C + D)i

3) Once again taking two complex numbers A + Bi and C + Di, we can take the difference of the two numbers, A + Bi - (C + Di) = (A - C) + (B - D)i

4) Finally we can multiply our two complex numbers together, using FOIL and the fact that $i^{2}=-1$ to find that $(A+Bi)(C+Di)=AC+ADi+BCi+BDi^{2}=(AC-BD)+(AD+BC)$ For example, $(3+4i)(2-i)=6-3i+8i-4i^{2}=(6+4)+(8-3)i=10+5i$

Conjugate

Since the Complex Numbers have similar properties to the Real Numbers, we would like to know if Complex Numbers have multiplicative inverses, that is if $A+Bi$ is a Complex Number is there another Complex Number C + Di such that (A + Bi)(C + Di) = 1. We know this holds for non-zero Real Numbers, by taking the reciprocal. For example, the multiplicative inverse of 5 is ${\frac {1}{5}}$

Before we can answer this question we are going to define the Conjugate:

If $z=A+Bi$ is a Complex Number, then its conjugate, denoted ${\overline {z}}$ is defined as $A-Bi$

Now we have the property that if $z=A+Bi$ then $z{\overline {z}}=A^{2}+B^{2}$

Now we can look at some properties of the Conjugate:

Given complex numbers z and w:

1)$({\overline {\overline {z}}})=z$

2) ${\overline {z+w}}={\overline {z}}+{\overline {w}}$

3) ${\overline {z\cdot w}}={\overline {z}}{\overline {w}}$

Now we are ready to find the inverse of any non-zero complex number A + Bi, that is find a complex number C + Di such that (A + Bi)(C + Di) = 1.

By the properties above, we know that $(A+Bi)(A-Bi)=A^{2}+B^{2}$ So given any complex number A + Bi, the inverse is ${\frac {A-Bi}{A^{2}+B^{2}}}$. Another way to think of this is ${\frac {1}{A+Bi}}={\frac {A-Bi}{A^{2}+B^{2}}}$.

Example: Find the inverse of 3 + 4i.

First we find the value of $A^{2}+B^{2}$ for A = 3 and B = 4. So $A^{2}+B^{2}=3^{2}+4^{2}=9+16=25$. Now the inverse of 3 + 4i is ${\frac {3-4i}{25}}$

Another common question that requires knowledge of the inverse is to simplify an expression of the form ${\frac {A+Bi}{C+Di}}$.

Example:

Simplify ${\frac {1+2i}{5+3i}}$.

From the discussion about inverses above, we can rewrite this expression as $(1+2i)\cdot {\frac {1}{5+3i}}$, and the inverse of the second term is ${\frac {5-3i}{5^{2}+3^{2}}}={\frac {5-3i}{34}}$. Putting this all together our expression has turned into $(1+2i)\cdot {\frac {5-3i}{34}}$. To finish the problem we just need to use FOIL and simplify. $(1+2i)\cdot {\frac {5-3i}{34}}={\frac {5+10i-3i-6i^{2}}{34}}={\frac {11+7i}{34}}$

Powers if i

Since the number i is so integral to defining the complex numbers, it is useful to know what happens when you raise i to different powers. Some powers of i are provided here:

$i^{1}=i$

$i^{2}=-1$

$i^{3}=-i$

$i^{4}=1$.

We can notice that each time we multiply i by itself 4 times we get 1. So if we want to know what i to any integer power is, we just have to find the remainder of the power when divided by 4 and use the powers of i that we see above.

For example, let's say we want to find the value of $i^{49}$. We divide 49 by 4 to get a reminder of 1. Looking at our values above, we can conclude that $i^{49}=i$

Revisiting the Quadratic Formula

On this page we learned how to solve quadratic equations using the quadratic formula. Now that we know about Complex Numbers we can classify all possible solutions to a quadratic equation and give it some geometric meaning. By geometric meaning, we can describe the behavior of the graph of the function.

We start by reminding the reader of the quadratic formula: Given a quadratic equation $ax^{2}+bx+c=0$ where a, b, and c are real numbers with $a\neq 0$, the solutions are given by $x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$, that is the solutions are $x={\frac {b^{2}+{\sqrt {b^{2}-4ac}}}{2a}}{\text{ and }}x={\frac {b^{2}-{\sqrt {b^{2}-4ac}}}{2a}}$.

The behavior of the graph is completely determined by the value of $b^{2}-4ac$ and we start by explaining how the value of $b^{2}-4ac$ affects the zeros of the quadratic equation $ax^{2}+bx+c=0$

1) If $b^{2}-4ac>0$ the equation $ax^{2}+bx+c$ has two distinct real roots.

2) If $b^{2}-4ac=0$ the equation $ax^{2}+bx+c$ has a single real double root, that is $ax^{2}+bx+c$ factors into the form $(x+d)^{2}$ for some real number d.

3) If $b^{2}-4ac<0$ the equation $ax^{2}+bx+c=0$ has two distinct complex roots, $z{\text{ and }}{\overline {z}}$

Geometrically each of these situations tells us the following about the graph of $ax^{2}+bx+c$

1) In this case the graph will cross the x-axis twice.

2) The graph will touch, and bounce away from the x-axis, staying on the same side. You can think of this like bouncing a tennis ball against a line on a tennis court, where the line is the x-axis. As long as you do not throw the tennis ball hard enough to create a crater the tennis ball has to stay "above" the line.

3) The graph will not even touch the x-axis, always staying on the same side.

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